A Fix on the Mistake

So here is one proof for the question from the previous post.  In case you forgot I shall now restate it:

Assume that in a metric space you are given 2 closed disjoint sets and .  Prove that if one of them is compact then with and .

proof: Without loss of generality (WLOG) assume that is compact.  Then define as .  Otherwise, define it to be the infimum of the distance from the point in to a point in the set .  Now is continuous and thus attains its infimum.  But this must be positive, because if it were not we would have contradicted the fact that the sets were disjoint and closed.

This proof works, but I would also like to find one that relies on a set being compact if any open cover has a finite subcover.  Sure it might be more work than necessary, but I feel that it should be good exercise.